Syllabus Edition

First teaching 2023

First exams 2025

|

The Mole & the Avogadro Constant (CIE AS Chemistry)

Revision Note

Test yourself
Richard

Author

Richard

Last updated

The Mole & the Avogadro Constant

  • The Avogadro constant (NA or L) is the number of particles equivalent to the relative atomic mass or molecular mass of a substance
    • The Avogadro constant applies to atoms, molecules, ions and electrons

  • The value of NA is 6.02 x 1023 g mol-1
  • The mass of a substance with this number of particles is called a mole (mol)
    • The mass of a substance containing the same number of fundamental units as there are atoms in exactly 12.00 g of 12C

  • One mole of any element is equal to the relative atomic mass of that element in grams
    • One mole of carbon, that is if you had 6.02 x 1023 atoms of carbon in your hand, would have a mass of 12 g
    • One mole of water would have a mass of (2 x 1 + 16) = 18 g

Worked example

Determine the number of atoms, molecules and the relative mass of 1 mole of:

  1. Na
  2. H2 
  3. NaCl

Answer 1

  • The relative atomic mass of Na is 23.0
  • Therefore, 1 mol of Na has a mass of 23.0 g mol-1
  • 1 mol of Na will contain 6.02 x 1023 atoms of Na (Avogadro’s constant)

Answer 2

  • The relative atomic mass of H is 1.0
  • Since there are 2 H atoms in H2, the mass of 1 mol of H2 is (2 x 1.0) = 2.0 g mol-1
  • 1 mol of H2 will contain 6.02 x 1023 molecules of H2
  • Since there are 2 H atoms in H2, 1 mol of H2 will contain 2 x 6.02 x 1023  = 1.204 x 1024 H atoms

Answer 3

  • The relative atomic mass of Na and Cl is 23.0 and 35.5 respectively
  • Therefore, 1 mol of NaCl has a mass of (23.0 + 35.5) = 58.5 g mol-1
  • 1 mol of NaCl will contain 6.02 x 1023 molecules of NaCl
  • Since there is one Na and one Cl atom in NaCl, 1 mol of NaCl will contain 2 x 6.02 x 1023  = 1.204 x 1024 atoms in total

1 mole of Number of atoms Number of molecules Relative mass
(g mol-1)
Na 6.02 x 1023 - 23.0
H2 1.204 x 1024  6.02 x 1023 2.0
NaCl 1.204 x 1024  6.02 x 1023 58.5

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Richard

Author: Richard

Expertise: Chemistry

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.